Let \(0\ne x\) be an element of centre of \(A\). Then \(I := \{ xy | y\in A\} \) is a two-sided-ideal of \(A\). Since \(0\ne x\in I\), we have that \(I = A\). Therefore \(1 \in I\), hence \(x\) is invertible.

If \(R\) is simple, then the \(\ker f\) is either \(\{ 0\} \) or \(R\). The former case implies that \(f\) is injective while the latter case implies that \(S\) is the trivial ring. Conversely, let \(I\subseteq R\) be a two-sided-ideal. Consider \(\pi : R \to {}^{R}/_{I}\), either \(\pi \) is injective implying that \(I = \{ 0\} \) or that \({}^{R}/_{I}\) is the trivial ring implying that \(I = R\).

By lemma 1.1.6, \(f\)is injective. Then \(\dim _{K}\operatorname{im}f = \dim _{K} B - \dim _{K}\ker f = \dim _{K}B\) meaning that \(f\) is surjective.

Assume \(A\) and \(B\) are central algebras, then by corollary 3.1.7 \(Z\left(A\otimes _{R}B\right)=Z\left(A\right)\otimes _{R}Z\left(B\right)=R\otimes _{R}R=R\).

By lemma 1.1.9, we need to prove \(A\otimes _{K}B\) is a simple ring. Denote \(f\) as the map \(A\to A\otimes _{K}B\). It is sufficient to prove that for every two-sided-ideal \(I\subseteq A\otimes _{K}B\), we have \(I = \left\langle f\left(f^{-1}\left(I\right)\right)\right\rangle \). Indeed, since \(A\) is simple \(f^{-1}\left(I\right)\) is either \(\left\{ 0\right\} \) or \(A\), if it is \(\left\{ 0\right\} \), then \(I=\left\{ 0\right\} \); if it is \(A\), then \(I\) is \(A\) as well.

We will prove that \(I \le \left\langle f\left(f^{-1}\left(I\right)\right)\right\rangle \), the other direction is straightforward. Without loss of generality assume \(I\ne \left\{ 0\right\} \). Let \(\mathcal{A}\) be an arbitrary basis of \(A\), by lemma 3.1.1, we see that every element \(x \in A\otimes _{K}B\) can be written as \(\sum _{i=0}^{n}\mathcal{A}_{i}\otimes b_{i}\) for some natrual number \(n\) and some choice of \(b_{i}\in B\) and \(\mathcal{A}_{i}\in \mathcal{A}\). Since \(I\) is not empty, we see there exists a non-zero element \(\omega \in I\) such that its expansion \(\sum _{i=0}^{n}\mathcal{A}_{i}\otimes b_{i}\) has the minimal \(n\). In particular, all \(b_{i}\) are non-zero and \(n\ne 0\). We have \(\omega =\mathcal{A}_{0}\otimes b_{0}+\sum _{i=1}^{n}\mathcal{A}_{i}\otimes b_{i}\). Since \(B\) is simple, \(1 \in B = \left\langle \langle b_{0} \right\rangle \); hence we write \(1\in \sum _{j=0}^{m}x_{i}b_{0}y_{i}\) for some \(x_{i},y_{i}\in B\). Define \(\Omega := \sum _{j=0}^{m}(1\otimes x_{i})\omega (1\otimes y_{i})\) which is also in \(I\). We write

For every \(\beta \in B\), we have that \(\left(1\otimes \beta \right)\Omega - \Omega \left(1\otimes \beta \right)\) is in I and is equal to

which is an expansion of \(n-1\) terms, thus \(\left(1\otimes \beta \right)\Omega -\Omega \left(1\otimes \beta \right)\) must be \(0\). Hence we conclude that for all \(i=1,\dots ,n\), \(\sum _{j=0}^{m}x_{j}b_{i}y_{j}\in Z\left(B\right)=K\). Hence for all \(i=1,\dots ,n\), we find a \(\kappa _{i}\in K\) such that \(\kappa _{i}=\sum _{j=0}^{m}x_{j}b_{i}y_{j}\). Hence we can calculate \(\Omega \) as

From this, we note that \(\mathcal{A}_{0}+\sum _{i}^{n}\kappa _{i}\cdot \mathcal{A}_{i}\in f^{-1}\left(I\right)\); since \(A\) is simple, we immediately conclude that \(f^{-1}\left(I\right) = A\), once we know \(\mathcal{A}_{0}+\sum _{i=1}^{n}\kappa _{i}\cdot \mathcal{A}_{i}\) is not zero. If it is zero, by the fact that \(\mathcal{A}\) is a linearly independent set, we conclude that \(1,\kappa _{1},\dots ,\kappa _{n}\) are all zero; which is a contradiction. Since \(f^{-1}\left(I\right) = A\), we know \(\left\langle f\left(f^{-1}I\right)\right\rangle = A \otimes _{K} B\).

By theorem 1.1.10, \(L\otimes _{K} D\) is simple. Let \(x\in Z\left(L\otimes _{K}D\right)\), by corollary 3.1.7, we have \(x\in Z\left(L\right)\otimes Z\left(D\right)=Z\left(L\right)\). Without loss of generality, we can assume that \(x = l \otimes d\) is a pure tensor, then \(l \in Z\left(L\right)\) and \(d \in K\). Therefore \(x = d\cdot l \in L\).

By symmetry, we only prove that \(A\) is simple. If \(A\) or \(B\) is the trivial ring then \(A\otimes _{K} B\) is the trivial ring, a contradiction. Thus we assume both \(A\) and \(B\) are non-trivial. Suppose \(A\) is not simple, by lemma 1.1.6, there exists a non-trivial \(K\)-algebra \(A'\) and a \(K\)-algebra homomorphism \(f : A \to A'\) such that \(\ker f \not= \{ 0\} \). Let \(F : A \otimes _{K} B\to A'\otimes _{K} B\) be the base change of \(f\), then since \(A\otimes _{K} B\) is simple and \(A'\otimes B\) is non-trivial (\(A'\) is non-trivial and \(B\) is faithfully flat because \(B\) is free), we conclude that \(F\) is injective. Then we have that

is exact. Since \(B\) is faithfully flat as a \(K\)-module, tensorig with \(B\) reflects exact sequences, therefore

is exact as well. This is contradiction since \(f\) is not injective.

By lemma 3.4.11, we have that \(\dim _{K} D = \dim _{K}C_{D}(k)\cdot \dim _{K}k\). Hence it is sufficient to show that \(C_{D}(k) = k\). By the commutativity of \(k\), we have that \(k \le C_{D}(k)\). Suppose \(k \ne C_{D}(k)\): let \(a \in C_{D}(k)\) that is not in \(k\). We see that \(L := k(a)\) is another subalgebra of \(D\) that is strictly larger than \(k\); a contradiction. Therefore \(k = C_{D}(k)\) and the theorem is proved.

We prove the following:

• “1. implies 2.”: this is lemma 3.4.11.

• “2. implies 1.”: Since \(L\) is commutative, we always have \(L \subseteq C_{A}(L)\). Hence we only need to show \(\dim _{K}L = \dim _{K}C_{A}(L)\). This is because by lemma 3.4.11, we have that \(\dim _{K}A = \dim _{K}L\cdot \dim _{K}C_{A}(L)\) and by 2. we have \(\dim _{K}L\cdot \dim _{K}C_{A}(L)= \dim _{K}L\cdot \dim _{K}L\).

• “2. implies 3.”: Since 2. implies 1., we assume \(L = C_{A}(L)\), therefore all we need is to prove \(L' \subseteq C_{A}(L)\). Let \(x \in L'\) and \(y \in L \subseteq L'\), we need to show \(xy = yx\) which is commutativity of \(L'\).

• “3. implies 1.”: By commutativity of \(L\), we always have \(L \subseteq C_{A}(L)\). For the other direction, suppose \(C_{A}(L)\not\subseteq L\), then there exists some \(a \in C_{A}(L)\) but not in \(L\). Consider \(L' = L(a)\), by 3., we have \(L' = L\) which is a contradiction.